6th Edition 8.73

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305113590
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Joined: Fri Sep 28, 2018 12:28 am

6th Edition 8.73

Postby 305113590 » Fri Feb 01, 2019 10:52 pm

Use the bond enthalpies in Tables 8.6 and 8.7 to estimate the reaction enthalpy for: CH4(g)+4Cl2(g)--->CCl4(g)+4HCl(g).

Can someone break down this step for me? I did 4(412)+4(0) - 4(338)- 4(431)=-1428 kJ/mol.
However, when I accounted for Cl2(g) being 242kJ/mol, I got the textbook answer of -460 kJ/mol. I thought Cl2 was 0.

Matthew Tran 1H
Posts: 165
Joined: Fri Sep 28, 2018 12:16 am

Re: 6th Edition 8.73

Postby Matthew Tran 1H » Fri Feb 01, 2019 11:55 pm

You're confusing standard enthalpy of formation and bond enthalpy. The standard enthalpy of formation of Cl2(g) is 0 because that is chlorine's standard state; however, we know that bond enthalpies are never 0 because breaking bonds require energy and forming bonds release energy. Therefore, you would need energy to break the Cl--Cl bond.

Aili Ye 4L
Posts: 58
Joined: Fri Sep 28, 2018 12:16 am

Re: 6th Edition 8.73

Postby Aili Ye 4L » Sat Feb 02, 2019 4:42 pm

As the other person mentioned, standard enthalpy of formation and bond enthalpy is different. Bond enthalpy cannot be zero since all bonds require energy to break.


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